Let group G= Z_6 × S_3 and H, K be 2 subgroups of G such that |H|=4 and |K|=9
I know |G|=2^2 * 3^2
What is the normality between the groups
(i) H & G
(ii) K & G
I know that a group is normally iffy the no of P-SSG of it is unique.
What does unique here means?
How do we find if H&G and K&G are normal?
A subgroup H of a group G is normal if and only if for every g in G, gHg^-1 = H. In other words, if conjugating any element of H by any element of G gives us an element that remains in H, then H is a normal subgroup of G.
To determine if H&G and K&G are normal, we need to examine if every element of G conjugates any element in H and K, respectively, without taking them outside of H and K.
For (i) H&G: Since H has only 4 elements, it is a cyclic subgroup of G generated by (a, b), where a is an element of Z_6 and b is an element of S_3. We can list all the elements of H as {(0, id), (3, id), (1, (1 2)), (4, (1 2))}, where id is the identity element of S_3.
Now, for any (x, y) in G, we need to check if it conjugates any of these 4 elements out of H. Let's suppose (x, y) does take some element outside of H. Then (x, y)(a, b)(x, y)^-1 = (xa(x^-1), yby^-1) must not lie in H. But since G is the direct product of Z_6 and S_3, this means xa(x^-1) must be in Z_6, and yby^-1 must be in S_3.
Since H is a subgroup of G, xa(x^-1) and yby^-1 must be elements of H as well. Therefore, we can conclude that H is a normal subgroup of G because for any (x, y) in G, conjugating any element of H by (x, y) gives us an element that remains in H.
For (ii) K&G: Since |K| = 9 is not a factor of |G| = 2^2 * 3^2, K cannot be a subgroup of G. Therefore, K&G is not even a well-defined subset of G, and we cannot determine if it is a normal subgroup.
Order is G is 36, how can K not be a subgroup of it when order of K is 9?
I apologize for the confusion in my previous response. You are correct that if the order of a subgroup K of a group G is a factor of the order of G, then K must be a subgroup of G.
In this case, since the order of K is 9 and the order of G is 36, K is indeed a subgroup of G. To determine if K is a normal subgroup of G, we need to check if for every g in G, gKg^-1 = K.
Let's first count the number of elements in K. By Lagrange's theorem, the order of every subgroup divides the order of the group. Since the order of K is 9 and the order of G is 36, the only possibilities for the order of K are 1, 3, and 9. K cannot have order 1 since it contains the identity element, and it cannot have order 3 since 3 does not divide 9. Therefore, K must have order 9 and G/K must have order 4 by Lagrange's theorem.
Since the order of G/K is 4, it must be isomorphic to either Z_4 or Z_2 × Z_2. Let's consider each case separately:
Case 1: G/K is isomorphic to Z_4.
In this case, there exists some k in K such that K = {0K, kK, 2kK, 3kK}. Let's suppose that gK = aK is some other coset in G/K. Then we can write g = ak + x, where x is some element of K. Now, for any h in K, we have
ghg^-1 = (ak + x)h(k^-1a^-1 + x') = akhk^-1a^-1 + xhx'.
Since K is a subgroup of G, hk^-1 is also in K. Therefore, ghg^-1 belongs to the coset aK if and only if xhx' is in K. But since K is a subgroup of index 4 in G, there are only 4 cosets, and each element of G belongs to one of these cosets. Therefore, we can conclude that K is a normal subgroup of G.
Case 2: G/K is isomorphic to Z_2 × Z_2.
In this case, there exists some k in K such that K = {(0,0)K, (0,1)K, (1,0)K, (1,1)K}. Let's suppose that gK = (a,b)K is some other coset in G/K. Then we can write g = (ak, bk) + (x,y), where (x,y) is some element of K. Now, for any h in K, we have
ghg^-1 = (ak, bk)(h,0)(k^-1a^-1, k^-1b^-1)(x,y)(kak^-1, kbk^-1) = (ak, bk)h(k^-1a, k^-1b)(kxk^-1 + akxk^-1 - ax, bxyk^-1)
Since K is a subgroup of G, hk^-1 is also in K. Therefore, ghg^-1 belongs to the coset (a,b)K if and only if kxk^-1 + akxk^-1 - ax is 0 (mod 2) and bxyk^-1 is 0 (mod 2). But since K is a subgroup of index 4 in G, there are only 4 cosets, and each element of G belongs to one of these cosets. Therefore, we can conclude that K is a normal subgroup of G.
Hence, we have shown that K is a normal subgroup of G in both cases, and therefore, K is a normal subgroup of G.
To determine if H and G are normal, we need to check if H is normal in G, i.e., if every element of G conjugates H back into H. Similarly, to determine if K and G are normal, we need to check if K is normal in G.
For H and G:
To find whether H is normal in G, we need to check if every element in G conjugates H back into H.
To do this, let's consider the elements of G. G = Z6 × S3, and its elements are pairs (a, b), where a is from Z6 (integers modulo 6), and b is from S3 (permutations of 3 elements).
So, an element in G can be written as (a, σ), where a is an integer from 0 to 5, and σ is a permutation from S3.
To check if H is normal in G, we need to check if for every element (a, σ) in G, (a, σ)H(a, σ)^(-1) = H.
Let's consider the elements of G one by one and see if they conjugate H back into H.
1. (0, e): Here, e is the identity permutation in S3. (0, e)H(0, e)^(-1) = H, where the exponentiation is the operation of conjugation.
2. (1, e): Similarly, (1, e)H(1, e)^(-1) = H.
3. (2, e): Again, (2, e)H(2, e)^(-1) = H.
4. (3, e): Likewise, (3, e)H(3, e)^(-1) = H.
5. (4, e): (4, e)H(4, e)^(-1) = H.
6. (5, e): (5, e)H(5, e)^(-1) = H.
7. (0, (12)): Here, (12) is the permutation (1 2). (0, (12))H(0, (12))^(-1) = H.
8. (0, (23)): (0, (23))H(0, (23))^(-1) = H.
9. (0, (13)): (0, (13))H(0, (13))^(-1) = H.
10. (0, (123)): (0, (123))H(0, (123))^(-1) = H.
11. (0, (132)): (0, (132))H(0, (132))^(-1) = H.
12. (1, (12)): (1, (12))H(1, (12))^(-1) = H.
...
The calculations can become quite tedious, as you need to check all possible combinations for each element. But if you do the calculations, and if all the conjugates of H yield elements in H, then H is normal in G.
You can follow the same process to determine if K is normal in G.
Now, coming to your question about "unique," in the context of normality, it means that there is only one possible subgroup in G that satisfies the condition of being normal. If a subgroup is normal, there cannot be another subgroup in the group that satisfies the same condition simultaneously.
So, to summarize, to determine if H and G, and K and G are normal, you need to check if, for every element in G, the conjugate of H or K respectively returns elements that are in H or K.